Multiplication of two vectors \(x, y \in \wedge ^1 \mathbb {R}^m \subset \mathbb {G}_m\) , such that \(x^2=y^2=1,\)  defines a multivector
\(xy = \cos (\theta ) - \sin (\theta )(x\wedge y),\) 
where \(\theta \)  is the angle between \(x\)  and \(y\) .
It is easy to see that \( (xy) (xy)^* = xyyx =1\) . The classical property of this geometric algebra is that the conjugation by such multivector \(xy\)  represents rotation in the plane \(x \wedge y\)  with respect to angle \(2 \theta \) , see Lemma 4.2 in [1]} for more details.
Using standard trigonometric formulas
we can verify this by straightforward calculation
\(\begin{split}R_{xy} &= \overline{1 + y x} =\frac{1 + \cos (\theta ) +( x \wedge y) \sin (\theta )}{\sqrt{(1 +\cos (\theta ))^2+\sin ^2(\theta )}}=\frac{1 + \cos (\theta ) + (x \wedge y) \sin (\theta )}{\sqrt{2 +2\cos (\theta )}} \\&=\sqrt{ \frac{1 + \cos (\theta )}{2}} + ( x \wedge y) \sqrt{\frac{1-\cos ^2(\theta )}{2(1 +\cos (\theta ))}}=\sqrt{ \frac{1 + \cos (\theta )}{2}} + ( x \wedge y) \sqrt{\frac{1-\cos (\theta )}{2}} \\&= \cos (\frac{\theta }{2}) + ( x \wedge y) \sin (\frac{\theta }{2}).\end{split}\) 

Finally \((x \wedge y)^* \cdot x = x \wedge y \wedge x =0\)  and
\((x \wedge y)^* \cdot y = 0\)  hold, so we compute
\( xy (x \wedge y)^* yx = xyyx(x \wedge y)^* = (x \wedge y)^* \) 
which proves the statement.
